Question

If the bisectors of angles $\angle ABC$ and $\angle ACB$ of a triangle ABC meet at a point O, then prove that $\angle BOC={90}^o+\frac{1}{2}\angle A$.

Answer 1

Given :

A $△ABC$ such that the bisectors of $\angle ABC$ and $\angle ACB$ meet at a point O.

To prove :

$\angle BOC={90}^o+\frac{1}{2}\angle A$

Proof :

In $△BOC$,

$\angle 1+\angle 2+\angle BOC={180}^o$

In $△ABC$,

$\angle A+\angle B+\angle C={180}^o$

$\angle A+2\left(\angle 1\right)+2\left(\angle 2\right)={180}^o$

$\frac{\angle A}{2}+\angle 1+\angle 2={90}^o$

$\angle 1+\angle 2={90}^o-\frac{\angle A}{2}$

Therefore,

${90}^o-\frac{\angle A}{2}+\angle BOC={180}^o$

$\angle BOC={90}^o+\frac{\angle A}{2}$